M12-13

Official Solution:


What is the equation of the line \(k\) that is perpendicular to line \(y = 2x\) and passes through point \((a, b)\)?

Since line \(k\) is perpendicular to \(y=2x\), its slope must be the negative reciprocal of the slope of \(y=2x\), which is \(2\). Therefore, the slope of line \(k\) is \(-\frac{1}{2}\). Using the formula for slope, which is the "rise over run" or change in \(y\) divided by change in \(x\), we can find the equation of line \(k\) passing through the point \((a,b)\) with slope \(-\frac{1}{2}\). Thus, we have the equation \(-\frac{1}{2}=\frac{y-b}{x-a}\), which simplifies to \(y=-\frac{x}{2}+\frac{a}{2}+b\).

(1) \(a = -b\).

If \(a=b=0\), the equation of \(k\) becomes \(y=-\frac{x}{2}+\frac{a}{2}+b=-\frac{x}{2}\), but if \(a = 1\), and \(b= -1\), the equation of \(k\) becomes \(y=-\frac{x}{2}+\frac{a}{2}+b=-\frac{x}{2}+\frac{1}{2}-1\). These are two different equations, so statement (1) alone is not sufficient.

(2) \(a - b = 1\).

If \(a=1\) and \(b=0\), the equation of \(k\) becomes \(y=-\frac{x}{2}+\frac{a}{2}+b=-\frac{x}{2}+\frac{a}{2}\), but if \(a = 0\), and \(b= -1\), the equation of \(k\) becomes \(y=-\frac{x}{2}+\frac{a}{2}+b=-\frac{x}{2}-1\). Again, these are two different equations, so statement (2) alone is not sufficient.

(1)+(2) We have two distinct linear equations \(a = -b\) and \(a - b = 1\), which we can solve for \(a\) and \(b\), and get the exact equation of line \(k\). Solving gives \(a=\frac{1}{2}\) and \(b=-\frac{1}{2}\). Therefore, the equation of line \(k\) is \(y=-\frac{x}{2}+\frac{a}{2}+b=-\frac{x}{2}+\frac{1}{4}-\frac{1}{2}=-\frac{x}{2}-\frac{1}{4}\). Sufficient.


Answer: C
_________________

Hi friends,

the line y=2x surely passes through the origin as the coordinates (0,0) satisfies the equation.

Now coming to the point, the solution says both the statements are required, but if you take the slope of second line as -1/2 then the equation for is

-1/2=(y-b)/(x-a)
or, 2y= -x +(a+b)
or, y=(-1/2)x+(a+b)/2

Now, applying statement 1 (a=-b)to this equation;

y=(-1/2)x+{a+(-a)}/2
or, y=(-1/2)x+ 0/2
or, y=(-1/2)x

So, statement 1 seems sufficient (contradiction with the actual answer)

Check the diagram below:



As you can see if a = 1 and b = -1 we have different line perpendicular to y = 2x than if a = 2 and b = -2.



_________________

Hi, Can anyone tell me what will be the final equation of the line once we know what a and b are?

From a=-b and a-b=1, we can get that a = 1/2 and b = -1/2. So, we know that k passes through the point (1/2, -1/2) and is perpendicular to line y = 2x (slope = 2)

The two lines are perpendicular if and only if the product of their slopes is -1, so m*2 = -1 and m = -1/2 (the slope of line k).

Finally, the equation of a straight line that passes through a point \(P_1(x_1, y_1)\) with a slope m is: \(y-y_1=m(x-x_1)\). Substitute: \(y-(-\frac{1}{2})=-\frac{1}{2}(x-\frac{1}{2})\) --> \(y = -\frac{x}{2} - \frac{1}{4}\).

Check below:




For Coordinate Geometry check:

24. Coordinate Geometry

• Theory
  Math: Coordinate Geometry
  Beauty of Coordinate geometry
  Dealing with Tangents on the GMAT
  
  • Questions
    DS Questions
    PS Questions

For more check:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.



_________________

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
_________________